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The flux  linked with secondary coil due to current I _1 = 2Amp  in primary coil is 4 \times 10 ^{-5}Wb

No. of turns in primary and secondary coil is 200 and 400 respectively . find the coeffeicient of mutual induction 

Option: 1

8 mH


Option: 2

4 mH


Option: 3

8 \times 10^{-4} H


Option: 4

4 \times 10 ^{-4}H


Answers (1)

best_answer

As we have learned

If the flux linked with one turn of coil is \Phi, then flux linked with N number of turns of coil will be N \Phi.

So

Co efficient of mutual induction -

(\phi _T)_2=N_{2}\phi_{2} \, \alpha \, I_{1}\Rightarrow \\ (\phi _T)_2=N_{2}\phi_{2} =MI_{1}
 

- wherein

N_{1}= Number of turns in primary

N_{2}= Number of turns in secondary

I_{1}= current through primary.

 

 Using

N_2\phi _2 = M I_1

M= N_2\phi _2 / I_1

M= \frac{400\times 4\times 10 ^{-5}}{2}

M=8 mH

 

 

 

 

Posted by

Rakesh

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